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我在 Yii 框架中工作,我有一个问题。我有一张带便利设施的桌子,我想找到所有东西。

$amenities = RoomTypeAmenity::model()->findAll(); 这个命令工作正常。

然后我想找到这个房间有什么设施,然后输入 $amenities_room = RoomTypeAmenity::model()->with('idRoomTypeAmenity')->findAll(); 这个命令工作正常。但是当打印设施时,我想检查现有的设施。

    foreach ($amenities as $amenity) { ?>
        <input type="checkbox" class="css-checkbox" id="facility_<?php echo $amenity->id_room_type_amenity; ?>">
        <label class="css-label" for="facility_<?php echo $amenity->id_room_type_amenity; ?>"><?php echo $amenity->name; ?></label>

        <?php  }?>
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1 回答 1

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使用if else条件里面for-loop。检查当前 amenity_room 是否存在当前 amenity。如果退出,checkedcheckbox.

foreach ($amenities as $amenity) { ?>
     if($amenity->id_room_type_amenity!=null){
         <input type="checkbox" class="css-checkbox" id="facility_<?php echo $amenity->id_room_type_amenity; ?>" checked>
         <label class="css-label" for="facility_<?php echo $amenity->id_room_type_amenity; ?>"><?php echo $amenity->name; ?></label>
     } else {
         <input type="checkbox" class="css-checkbox" id="facility_<?php echo $amenity->id_room_type_amenity; ?>">
         <label class="css-label" for="facility_<?php echo $amenity->id_room_type_amenity; ?>"><?php echo $amenity->name; ?></label>
     }

<?php  }?>

希望这可以帮助!

于 2013-05-15T03:52:33.207 回答